The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at

Question

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at (A) (9R/400) cm-1 (B) (7R/144) cm-1 (C) (3R/4) cm-1 (D) (5R/36) cm-1

Tardigrade Admin · Accepted Answer

For Balmer n1 = 2 and n2 = 3; v = R ((1/22) - (1/32)) = (5R/36) cm-1

Q. The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at

$\frac{9 R}{400} c m^{- 1}$

$\frac{7 R}{144} c m^{- 1}$

$\frac{3 R}{4} c m^{- 1}$

$\frac{5 R}{36} c m^{- 1}$

For Balmer $n_{1} = 2$ and $n_{2} = 3; v$
$= R (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = \frac{5 R}{36} c m^{- 1}$

Q. The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at

4009R​cm−1

1447R​cm−1

43R​cm−1

365R​cm−1

For Balmer n1​=2 and n2​=3;v =R(221​−321​)=365R​cm−1

$\frac{9 R}{400} c m^{- 1}$

$\frac{7 R}{144} c m^{- 1}$

$\frac{3 R}{4} c m^{- 1}$

$\frac{5 R}{36} c m^{- 1}$

For Balmer $n_{1} = 2$ and $n_{2} = 3; v$
$= R (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = \frac{5 R}{36} c m^{- 1}$